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Let $R$ be a commutative ring, with 1.

Prove that if $R$ is reduced, then $R$ is integrally closed in $R[X]$, i.e. $R \subset R[X]$ is an integral extension of rings.

I found this problem in many introductory courses, but I simply can't solve it.

Thanks in advance.

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Hmm, something that may be hampering you is that you might be confusing two things. Let $R\subseteq S$ be a ring extension and write the integral closure of $R$ as $\overline{R}$. Then $R$ is called integrally closed if $R=\overline{R}$. The extension is called integral if $S=\overline{R}$. –  rschwieb Jun 3 '12 at 18:11
    
I should have qualified "write the integral closure of $R$ in $S$ as $\overline{R}$". –  rschwieb Jun 3 '12 at 18:23
    
I translated the problem as faithful as I could (I'm Romanian, it's a problem our professor gave us). What I think I have to prove is that there exists a monic poly $g \in R[X]$ s.t. $\forall 0 \neq f \in R[X], \ g \circ f=0$, i.e. $f^n+\displaystyle\sum_{i=0}^{n-1} a_i \cdot f^i=0$, where $a_i \in R$ are the coeffs of $g$. Right? If so, what next? –  AdrianM Jun 3 '12 at 18:33
    
I'm 99% certain that the goal is to show that $\overline{R}=R$. That means that no polynomial with degree>0 is integral over $R$. –  rschwieb Jun 3 '12 at 18:38
    
The strategy you gave in the last comment is way off track. It looks like you are trying to show that there is a monic polynomial that has everything in $R[x]$ as a root. That should already be setting off some alarms :) –  rschwieb Jun 3 '12 at 18:47

1 Answer 1

up vote 3 down vote accepted

The task is to show that the only elements of $R[x]$ integral over $R$ are those already in $R$.

Here's a start:

Suppose $p(x)\in R[x]\setminus{R}$ is integral over $R$, and say the degree of $p(x)$ is $n>0$.

Key observation: Since $R$ is reduced, the $i$'th power of $p(x)$ has degree $n*i$.

We have that $p(x)$ satisfies some monic polynomial over $R$, say of degree $k$. Rewrite that equation as $(p(x))^k=$ (sum of lower powers of $p(x)$ with coefficients from $R$).

Can you see why this results in a contradiction?

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...because of equality of degrees? $deg(p^k)=n\cdot k$, and deg of RHS cannot exceed $n \cdot (k-1)$..? –  AdrianM Jun 3 '12 at 18:49
    
Right :) Good luck with your studies! –  rschwieb Jun 3 '12 at 18:51
    
Thanks a million! I really need a lot of luck, since I'm getting ready for a really tough exam in ComAlg. But my personal interest is NONComAlg, which seems (to me) something from another world. –  AdrianM Jun 3 '12 at 18:58
    
@AdrianM I'm a noncommutative algebraist too, and I wish my commutative algebra were better. The commutative algebraists are able to get really deep results because their rings are so nice. –  rschwieb Jun 3 '12 at 18:59
    
I have an organic repulsion to commutative algebra, since my university course is very computational (even) and I like a lot more working with categories or homological algebra. You are right, though, that commutative results are deeper, but let's say it's just not my cup of tea. –  AdrianM Jun 3 '12 at 19:06

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